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- Thread starter goody
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- Aug 30, 2012

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\(\displaystyle 2^x \approx 1 + (x ~ ln(2) ) + \dfrac{1}{2} (x ~ ln(2) )^2 + \dfrac{1}{6} (x ~ ln(2) )^3 + \text{ ...}\)

\(\displaystyle 2^{sin(x)} \approx 1 + (x ~ ln(2) ) + \dfrac{1}{2} (x ~ ln(2) )^2 + \dfrac{1}{6} x^3 ( (ln(2)^3 - ln(2) ) + \text{ ...}\)

\(\displaystyle cos(x) \approx 1 - \dfrac{1}{2} x^2 + \text{ ...}\)

(These are MacLaurin series.)

So

\(\displaystyle \dfrac{ 2^x - 2^{sin(x)} }{x (1 - cos(x) ) } \approx \dfrac{ \dfrac{1}{6} x^3 ln(2) }{x \left ( 1 - \left ( 1 - \dfrac{1}{2} x^2 \right ) \right ) } = \dfrac{1}{3} ln(2)\)

Ugly, but doable. (And yes, I did the \(\displaystyle 2^{sin(x)}\) series myself. I only checked it with W|A.)

-Dan

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